Answer :

ujalakhan18

Answer:

[tex]\huge\boxed{f^{-1}(x) = (x-3)^2+2}[/tex]

Step-by-step explanation:

[tex]f(x) = \sqrt{x-2} + 3[/tex]

Replace y = f(x)

[tex]y = \sqrt{x-2} + 3[/tex]

Exchange x and y

[tex]x = \sqrt{y-2}+3[/tex]

Solve for y

[tex]x = \sqrt{y-2}+3[/tex]

Subtracting both sides by 3

[tex]x - 3 = \sqrt{y-2}[/tex]

Taking square on both sides

[tex](x-3)^2 = y -2[/tex]

Adding 2 to both sides

[tex]y = (x-3)^2+2[/tex]

Substitute y = [tex]f^{-1}(x)[/tex]

[tex]f^{-1}(x) = (x-3)^2+2[/tex]

TheAnimeGirl

Answer:

[tex] \boxed{ {f}^{ - 1} (x) = {(x - 3)}^{2} + 2}[/tex]

Option D is the correct option

Step-by-step explanation:

[tex] \mathsf{f(x) = \sqrt{x - 2} + 3}[/tex]

Replace f(x) with y

[tex] \mathsf{y = \sqrt{x - 2} + 3}[/tex]

Interchange variables

[tex] \mathsf{x = \sqrt{y - 2} + 3}[/tex]

[tex] \mathsf{{(x - 3)}^{2} = {( \sqrt{y - 2)} }^{2} }[/tex]

[tex] \mathsf{ {(x - 3)}^{2} = y - 2}[/tex]

[tex] \mathsf{ y = {(x - 3)}^{2} + 2}[/tex]

Replace y with f ⁻¹( x )

[tex] \mathsf{ {f}^{ - 1} (x) = {(x - 3)}^{2} + 2}[/tex]

Hope I helped!

Best regards!

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